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mathcalculus Group Title

Help: CONTINUITY and removable discontinuity: (attached below)

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    • one year ago
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  2. mathcalculus Group Title
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    @kropot72

    • one year ago
  3. Psymon Group Title
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    So you have 2 conditions to have a removable discontinuity. The first one is simply that the value must not be defined at that point, which is pretty obvious that is isn't because of the denominator. The second condition is basically the difference between whether or not you have an asymptote at your undefined point or simply a hole. The way to tell whether or not its an asymptote or a whole is if you factored your function, would the factor that is undefined cancel or remain? If the factor on bottom cancels, it is merely a hole, if it does not cancel then it is an asymptote and CANNOT be removed. So I would factor your numerator and see if the denominator cancels. If it does, then simply plug in 4 into your result and you'll get the value you're looking for.

    • one year ago
  4. mathcalculus Group Title
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    can we work on this together?

    • one year ago
  5. mathcalculus Group Title
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    i've tried doing this on my own but seem to be getting something wrong.

    • one year ago
  6. Psymon Group Title
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    Sure. So I would factor the numerator and see if x-4 cancels. If it cancels, you have a hole, if it does not cancel, you have an asymptote and it cannot be removed. You know how to factor the numerator, right?

    • one year ago
  7. mathcalculus Group Title
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    let me try.

    • one year ago
  8. Psymon Group Title
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    Alright, sure.

    • one year ago
  9. mathcalculus Group Title
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    i got 2(x^2+3x-25) @Psymon

    • one year ago
  10. Psymon Group Title
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    -28you mean? And after that, you still have to further factor the x^2 + 3x - 28.

    • one year ago
  11. Psymon Group Title
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    Find factors of -28 that will add up to positive 3.

    • one year ago
  12. mathcalculus Group Title
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    yes, ,-28

    • one year ago
  13. Psymon Group Title
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    Right. So When you further factor the numerator, you should have 2(x±__)(x±__). Its just remember how to factor that x^2 + 3x - 28 portion.

    • one year ago
  14. mathcalculus Group Title
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    2(x+7) (x-4)

    • one year ago
  15. Psymon Group Title
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    Exactly. In turn, that cancels out the denominator, which means the discontinuity is removable and not just an asymptote. Not only that, since the term that would make your function undefined disappeared, you can simply plug in 4 to find out the value you need.

    • one year ago
  16. mathcalculus Group Title
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    we're left with 2 (x+7)

    • one year ago
  17. mathcalculus Group Title
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    so we plug in 4?

    • one year ago
  18. Psymon Group Title
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    That's all ya do.

    • one year ago
  19. mathcalculus Group Title
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    does this apply to every problem with continuity?

    • one year ago
  20. Psymon Group Title
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    Yes. Most of the work is finding a way to eliminate the portion that would give you an undefined answer. Once you manage that, you just plug in the value that the limit is approaching. Of course there are times when you cannot remove the undefined portion, in which that means the limit does not exist and you are merely approaching an asymptote.

    • one year ago
  21. mathcalculus Group Title
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    thanks:)

    • one year ago
  22. Psymon Group Title
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    Sure :3

    • one year ago
  23. mathcalculus Group Title
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    ohhh can i practice a little, then ask you for a tricky problem like the one you just mentioned? it would really help.

    • one year ago
  24. Psymon Group Title
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    That works, I have no problem with that.

    • one year ago
  25. mathcalculus Group Title
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    what about a JUMP DISCONTINUITY? same thing?

    • one year ago
  26. Psymon Group Title
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    That's new terminology for me. I probably know what it is but have not heard it said that way. Could you give me an example?

    • one year ago
  27. mathcalculus Group Title
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    sure

    • one year ago
  28. mathcalculus Group Title
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    1 moment

    • one year ago
  29. mathcalculus Group Title
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    • one year ago
  30. Psymon Group Title
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    Ah, just one sided limits. We never really gave a name for it before, just that it had to do with one-sided, lol.

    • one year ago
  31. mathcalculus Group Title
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    oh so how do we begin with this one? sketch a graph?

    • one year ago
  32. mathcalculus Group Title
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    i can do graphs but i'd like to do it a simpler way since i won't have enough time when it comes to a final

    • one year ago
  33. mathcalculus Group Title
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    @Psymon

    • one year ago
  34. Psymon Group Title
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    I apologize, I guess I was making sure what jump kind of meant since I did not have a picture. But yes, all you really need to confirm is that when x = 6 on both functions that they have different limits. It is pretty obvious that they do. As far as the left and right part goes, you need to see which function to just check and see which function is approaching from the left and which one is approaching from the right. Since 8x-8 < 6, that must be the function approaching from the left. Therefore, f(6) of 8x-8 is the value you will use for limit from the left. Since the other function is the one that will becoming from the right, it's value at x = 6 will be the value you use for the right-sided limit. Basically, you just need to see which function to plug the value into.

    • one year ago
  35. Psymon Group Title
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    The delay was graphing and looking at so I knew exactly what to say and to get my brain to work, lol.

    • one year ago
  36. mathcalculus Group Title
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    it's okay, so how can we work this prob out?

    • one year ago
  37. Psymon Group Title
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    |dw:1375046904734:dw| I know you know how to graph, but I wanted the visual up before we move on. So you just want to see which function is from the left and which one is from the right. When it gives you a piece-wise function, it's pretty obvious which one is going to be coming from the left and which one from the right. Once you determine that, just plug in x = 6 into the graph that comes from the left and use that value for lim ->6 - Once you do that, plug x = 6 into the other function, since you know it's the one on the right, and find lim x-> 6 +

    • one year ago
  38. Psymon Group Title
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    A lot of the time you may have to determine the behavior on your own, but the piece-wise just gave you the behavior so you didn't have to worry about it. From there it was just picking the correct function for left and for right.

    • one year ago
  39. Psymon Group Title
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    In regards to removable or non-removable,it works the same way. if one of the functions has an asymptote at the jump point, then one of the sides or both may have a limit that DNE on that side, left or right. Maybe another example?

    • one year ago
  40. mathcalculus Group Title
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    so to this problem we HAVE to sketch only?

    • one year ago
  41. Psymon Group Title
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    In this case, no, because the piece-wise function tells us which function to use on the left and on the right. The sketching would only come in if you had maybe one function that wasn't a piece-wise. Like if the function was 1/[(2x^2 -3] For something like that you would have to test points on the left and right and see if they're going up to infinity, down to negative infinity, etc. Any of these piece-wise functions won't require s ketch, I just wanted to draw visual aid.

    • one year ago
  42. Psymon Group Title
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    Holy cow my typing and grammar are horrible today o.o

    • one year ago
  43. Psymon Group Title
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    Maybe if you have another jump problem we can look at it.

    • one year ago
  44. Psymon Group Title
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    I'm going to need to head out, I apologize. Good luck with the limits.

    • one year ago
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